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Suppose that $X_ n \sim \text {Gamma}\ (n\alpha , \lambda)$ for all $n \ge 1$, for fixed $\alpha,\lambda >0.$ Show that

$$\frac {1} {\sqrt n} \left (X_n - \frac {n \alpha} {\lambda} \right ) \implies Z\ ,$$

where $Z \sim \mathcal N (0,{\sigma}^2)$ for some $\sigma.$ Calculate $\sigma.$

I have constructed a sequence random variables $\{S_n \}$ in the following way $:$

$S_1=X_1,S_2=X_2-X_1,S_3=X_3-X_2, \cdots , S_n = X_n - X_{n-1} , \cdots.$ Then I observed that $\sum\limits_{k=1}^{n} S_k = X_n$ for all $n \ge 1$. Also I observed that $\Bbb E(S_n) = \frac {\alpha} {\lambda}$ and $\Bbb {Var} (S_n) = \frac {\alpha} {{\lambda}^2}$ for all $n \ge 1$. If moreover $S_n$'s can be shown to be independent then they are i.i.d. random variables with mean $\frac {\alpha} {\lambda}$ and variance $\frac {\alpha} {{\lambda}^2}$. Then by central limit theorem we can say that

$$\frac {1} {\sqrt n} \left (X_n - \frac {n \alpha} {\lambda} \right ) \implies Z$$ as $n \rightarrow \infty$ where $Z \sim \mathcal N (0,\frac {\alpha} {{\lambda}^2})$. Hence ${\sigma}^2 = \frac {\alpha} {{\lambda}^2}$ i.e. $\sigma =\frac {\sqrt \alpha} {\lambda}.$

But how do I prove that $S_i$'s are i.i.d. random variables keeping in mind the fact that $X_i$'s are independent random variables? Please help me in this regard.

Thank you very much.

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  • $\begingroup$ Regarding your last paragraph: the $X_i$ are not i.i.d., as their distributions depend on $i$. $\endgroup$ – Quaternion Dec 1 '18 at 8:29
  • $\begingroup$ Sorry I mean to say that they are independent. Thanks for pointing this out @Quaternion. I will edit my question soon. $\endgroup$ – Dbchatto67 Dec 1 '18 at 8:36
  • $\begingroup$ @Dbchatto67 I don't think your variance of $S_n$ is calculated correctly. Note that $\textrm{V}(S_n) = \textrm{V}(X_n-X_{n-1}) = \textrm{V}(X_n)+\textrm{V}(X_{n-1}) = \frac{\alpha n}{\lambda^2} + \frac{\alpha (n-1)}{\lambda^2}$ $\endgroup$ – Lundborg Dec 1 '18 at 9:06
  • $\begingroup$ In the statement of your problem you do not mention independence of the $X_i$... $\endgroup$ – saz Dec 1 '18 at 9:20
  • $\begingroup$ I'm not sure if introducing the $S_n$'s and using the CLT is a good idea. Have you tried directly computing the limit while using the Stirling approximation for the $\Gamma(\alpha n)$ term? $\endgroup$ – Quaternion Dec 1 '18 at 9:21
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This answer uses the the following fact.

If $X \sim \Gamma(\alpha,\lambda)$ and $Y \sim \Gamma(\beta,\lambda)$ are independent, then $X+Y \sim \Gamma(\alpha+\beta,\lambda)$.

Hints: Let $(Y_i)_{i \in \mathbb{N}}$ be a sequence of independent identically distributed random variables such that $Y_i \sim \Gamma(\alpha,\lambda)$.

  1. Show that $\tilde{X}_n := \sum_{i=1}^n Y_i$ satisfies $\tilde{X}_n \sim \Gamma(n \alpha,\lambda)$.
  2. Apply the central limit theorem to prove that $$\frac{1}{\sqrt{n}} \left( \tilde{X}_n - \frac{n \alpha}{\lambda} \right) \stackrel{d}{\to} Z$$ for $Z \sim N(0,\sigma^2)$ with $\sigma^2 = \text{var}(Y_1)$; here $\stackrel{d}{\to}$ denotes convergence in distribution.
  3. Use the fact that $\tilde{X}_n$ equals in distribution $X_n$ for each $n \in \mathbb{N}$ to conclude from Step 2 that $$\frac{1}{\sqrt{n}} \left( X_n - \frac{n \alpha}{\lambda} \right) \stackrel{d}{\to} Z$$
  4. Compute $\sigma^2 = \text{var}(Y_1)$ (...or look it up, e.g. on wikipedia).
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  • $\begingroup$ So $Z \sim \mathcal N \left (0, \frac {\alpha} {{\lambda}^2} \right )$. Then $\sigma = \frac {\sqrt \alpha} {\lambda}$. Am I right @saz? $\endgroup$ – Dbchatto67 Dec 1 '18 at 11:28
  • $\begingroup$ @Dbchatto67 Yes, that's right. $\endgroup$ – saz Dec 1 '18 at 11:50
  • $\begingroup$ Could you take a look at my last question? would really appreciate it $\endgroup$ – badatmath Dec 1 '18 at 14:10
  • $\begingroup$ @itry To which end? There is an answer and you accepted it... so where is the problem? $\endgroup$ – saz Dec 1 '18 at 17:23
  • $\begingroup$ I mean the one with the bounty about cramers theorem $\endgroup$ – badatmath Dec 1 '18 at 17:24

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